This question has been asked before, but I don\'t believe the answer was entirel

Question

This question has been asked before, but I don't believe the answer was entirely correct. Really think about number 1 with B and how e would be affected.

A parallel plate capacitor was connected to a DC Battery and Charged. The separation between the plates was then increased by a factor of two. Apply the following scenarios:

1.) Before the separation was Increased the Capacitor was disconnected from the battery.

2.) The Capacitor remained connected to the Battery while the separation was Increased.

For both scenarios show How the following quantities Changed..
a.) Capacitance?
b.) Charge on the Capacitor?
c.) Potential Difference between the Plates?
d.) Electric Field between the Plates?
e.) Energy Accumulated in the Capacitor..In particular has there been work done to move the plates and has there been energy drawn from the battery or did the battery receive energy from the work done on separating the plates?

Explanation / Answer

initial capacitance of capacitor   C = o A /d if seperation between plates increases by a factor 2   a) capacitance of the capacitor is                  C'   = o A /2d                      =   ( 1/2) ( o A /d )                      = C /2       capacitance decreases to half of initial capacitance b) charge on the capacitor                 Q ' = C ' V                       = C V / 2                       = Q / 2        charge on the capacitor c) potential difference between the plates                 V' = Q / C                    =    V    d)   Electric field between the plates is              E' =   V / d                   = ( 1/2 ) ( V / d )                   = E / 2                    

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